Question

A group of 30 people consists of 15 children, 10 men, and 5 women. Tom and Jerry are two of the men in the group.


Five people are picked at random without replacement.
a) Find the chance the first person picked is a man, given that the fourth and fifth people picked are children.
b) Find the chance that more than two women are picked
c) Find the chance that Tom and Jerry both get picked.

Answer 1

somebody please reply

Answer 2

a. Take 2 children out of the group and place them at the 4th and 5th position resp., how many men and how many people in the group do we have now?
b. Could be (3 women, 2 men) or (3 women, 2 children) or (3 women, 1 man, 1 child) or (4 women, 1 man) or (4 women, 1 child) or (5 women), just add up the probabilities.
c. Take 2 men named Tom and Jerry out of the group, now how many men and how many people in the group do we have?

Answer 3

a)\[P(first\ 3\ have 1\ man)=\frac{2C1\times 26C2}{28C3}\]
\[P(first\ 3\ selected\ has\ man\ first)=\frac{2}{3\times 2}\]
\[P(first\ person\ selected\ is\ the\ 1\ man)=\frac{2C1\times 26C2}{28C3}\times \frac{2}{3\times 2}............(1)\]
\[P(first\ 3\ selected\ has\ 2\ men)=\frac{26}{28C3}\]
\[P(first\ person\ selected\ is\ one\ of\ the\ 2\ men)=\frac{4}{3\times 2}\]
\[P(first\ person\ selected\ is\ one\ of\ the\ 2\ men)=\frac{26}{28C3}\times \frac{4}{3\times 2}...........(2)\]
The chance the first person picked is a man, given that the fourth and fifth people picked are children is found by adding (1) and (2).